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Solve for the rational numbers x and y: \(2^{x+y} \cdot 3^{x-y} \cdot 6^{2x+2y}= 72.\)

 Jul 13, 2019
 #1
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Using Newton-Raphson method, it gives the following solution:

x = 1/2   and     y=1/2

 Jul 13, 2019
 #2
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\(2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,6^{2x+2y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,6^{2(x+y)}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,(6^2)^{x+y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,36^{x+y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,36^{x+y}\,\cdot\,3^{x-y}\ =\ 72\\~\\ (2\,\cdot\,36)^{x+y}\,\cdot\,3^{x-y}\ =\ 72\\~\\ 72^{x+y}\,\cdot\,3^{x-y}\ =\ 72\)

 

At this point we can see that the equation will definitely be satisfied if

 

x + y  =  1    and

x – y  =  0

 

So    x  =  1/2    and    y = 1/2    is definitely a solution.

 

But that doesn't necessarily mean it is the only solution.

 

I don't know how to prove that it is the only rational solution, but it does seem to be.

 

Maybe someone else can explain why that is.

 Jul 13, 2019
 #3
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LHS:

\(\displaystyle 2^{x+y}.3^{x-y}.6^{2x+2y}=2^{x+y}.3^{x-y}.(2\times3)^{2x+2y}=2^{x+y}.3^{x-y}.2^{2x+2y}.3^{2x+2y}\\=2^{3x+3y}.3^{3x+y}\)

 

RHS:

\(\displaystyle 72=2^{3}.3^{2}\)

 

It's an identity, so equating indices,

\(\displaystyle 3x+3y=3,\\3x+y=2.\)

 

Solve, x = 1/2, y = 1/2.

 

The solution is unique.

 Jul 14, 2019
 #4
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The expression p1Q1*p2Q2*....*pnQn for p1,....,pn distinct primes and Q1,....,Qn rational numbers will result in an integer if and only if Q1,...,Qn are all nonnegative integers

 Jul 14, 2019

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